(a) (i) Electric field due to an infinitely long thin straight wire with uniform linear charge density:Linear charge density $=\lambda$A point $P$ is considered at a distance $x$ where the electric field is to be determined. A cylindrical Gaussian surface of length $l$ and radius $x$ is considered with the wire as its axis. Magnitude of electric field at every point the curved surface of the cylinder is same and is directed outward making an angle $0^{\circ}$ with the direction of area vector . Electric fields at the two flat surfaces of the cylinder are zero since, the direction of electric field through those surfaces and the direction of area vector are perpendicular. So, total flux through the curved surface $=\phi=E\times 2\pi x l$ According to Gauss' law, $\phi=\;\frac{l\lambda}{\varepsilon_0}$ $\;\therefore\;E\times 2\pi x l\;=\;\frac{l\lambda}{\varepsilon_0}$ $\therefore\;E\;=\;\frac{1}{2\pi\varepsilon_0}\;\frac{\lambda}{x}$ (ii) Electric field produced by an infinitely long straight charged wire, $E=\;\frac{\lambda}{2\pi\varepsilon_0 r}$ This field provides the centripetal force to the revolving electron. $eE=\;\frac{mv^2}{r}$ or, $\;\;\;\frac{\lambda}{2\pi\varepsilon_0 r}=\;\frac{mv^2}{r}$ or, $mv^2=\;\frac{e\lambda}{2\pi\varepsilon_0}$ $\therefore$ Kinetic energy $(\text{KE})=\;\frac{1}{2}mv^2=\;\frac{e\lambda}{4\pi\varepsilon_0}$ (iii) The required equation, $\;\frac{1}{2}mv^2=\;\frac{e\lambda}{4\pi\varepsilon_0}$ It is the equation of a straight line passing through the origin. OR (b) (i) (1) Yes, Electric field will be zero at $\left(\;\frac{a}{2},0\right)$ point. At this point the magnitudes of both the electric fields are same in magnitude and are oppositely directed.(2) Electric potential will not be zero at any point since at every point the potential will be the summation of both the potentials and both charges are of similar nature.(ii) Significance of negative electrostatic potential energy:Negative electrostatic potential energy of a system of charges means work is to be done against the field to move the charges apart. Potential energy due to charges at A and B$\;U_1=\;\frac{kq_1q_2}{r}$$\;\text{Or,}\;\;U_1=\;\frac{9\times 10^9\times(-4)\times(4)\times 10^{-12}}{2}$$\therefore\;U_1=-72\times 10^{-3}\ \text{J}$Potential energy due to charges at $A$ and $C$$\;U_2=\;\frac{kq_1q_3}{r}$$\;\text{Or,}\;\;U_1=\;\frac{9\times 10^9\times(4)\times(2)\times 10^{-12}}{2}$$\therefore\;U_2=36\times 10^{-3}\ \text{J}$Potential energy due to charges at $B$ and $C$$U_3=\;\frac{kq_2q_3}{r}$ $\;\text{Or,}\;\;U_3=\;\frac{9\times 10^9\times(-4)\times(2)\times 10^{-12}}{2}$ $\therefore\;U_3=-36\times 10^{-3}\ \text{J}$ Net potential energy of the system of three charges. $\;U\;=U_1+U_2+U_3$ $\;\text{Or,}\;\;U\;=-72\times 10^{-3}\ \text{J}$ $+36\times 10^{-3}\ \text{J} -36\times 10^{-3}\ \text{J}$ $\therefore\;\;U=-72\times 10^{-3}\ \text{J}$